A line passing through the point A(-2, 0), touches the parabola P : $$y^2 = x - 2$$ at the point B in the first quadrant. The area of the region bounded by the line AB, parabola P and the x-axis, is :

The parabola is $$y^{2}=x-2 \; \; (1)$$.
A general point on it can be written in parametric form as $$\bigl(t^{2}+2,\;t\bigr)\;.$$

Equation of the tangent at the parametric point
For $$y^{2}=x-2$$ the standard slope-form of a tangent is obtained by differentiating: $$2y\dfrac{dy}{dx}=1\;\;\Longrightarrow\;\;\dfrac{dy}{dx}=\dfrac{1}{2y}\;.$$
At the point $$B\bigl(x_{1},y_{1}\bigr)=\bigl(t^{2}+2,\;t\bigr)$$ the slope is $$m=\dfrac{1}{2t}$$ and the tangent is $$y-t=\dfrac{1}{2t}\bigl(x-(t^{2}+2)\bigr)\;.\qquad -(2)$$

The tangent passes through the fixed point $$A(-2,0)$$, so $$(x,y)=(-2,0)$$ must satisfy (2):
$$0-t=\dfrac{1}{2t}\Bigl(-2-(t^{2}+2)\Bigr) \;\;\Longrightarrow\;\;-t=-\dfrac{t^{2}+4}{2t} \;\;\Longrightarrow\;\;2t^{2}=t^{2}+4 \;\;\Longrightarrow\;\;t^{2}=4\;.$$

Because point $$B$$ lies in the first quadrant, $$t=+2$$. Thus $$B(6,2)\;,\qquad m=\dfrac{1}{2t}=\dfrac14\;.$$

Equation of the required tangent (line AB)
Using point $$B(6,2)$$ and slope $$\dfrac14$$: $$y-2=\dfrac14(x-6)\quad\Longrightarrow\quad y=\dfrac{x}{4}+\dfrac12\;.\qquad -(3)$$

Sketch of the enclosed region
The boundary consists of • the segment of line (3) from $$A(-2,0)$$ to $$B(6,2)$$,
• the arc of the parabola (1) from $$D(2,0)$$ to $$B(6,2)$$, and
• the segment of the $$x$$-axis from $$A(-2,0)$$ to $$D(2,0)$$.

Area under the line from $$x=-2$$ to $$x=6$$
From (3), $$y=\dfrac{x}{4}+\dfrac12$$. $$\displaystyle A_{\text{line}}=\int_{-2}^{6}\Bigl(\dfrac{x}{4}+\dfrac12\Bigr)\,dx =\left[\dfrac{x^{2}}{8}+\dfrac{x}{2}\right]_{-2}^{6} =\left(\dfrac{36}{8}+3\right)-\left(\dfrac{4}{8}-1\right)=8\;.$$

Area under the parabola from $$x=2$$ to $$x=6$$
For the upper half of (1), $$y=\sqrt{x-2}$$. $$\displaystyle A_{\text{parab}}=\int_{2}^{6}\sqrt{x-2}\,dx =\frac23\bigl(x-2\bigr)^{3/2}\Bigl|_{2}^{6} =\frac23\,(4)^{3/2}=\,\frac{16}{3}\;.$$

Required area
Over $$[-2,2]$$ the region is the strip between the line and the $$x$$-axis; over $$[2,6]$$ it lies between the line and the parabola. Hence $$\displaystyle A=A_{\text{line}}-A_{\text{parab}} =8-\frac{16}{3}=\,\frac{8}{3}\;.$$

The area bounded by the line $$AB$$, the parabola $$y^{2}=x-2$$ and the $$x$$-axis equals $$\displaystyle \frac{8}{3}$$.
Therefore, the correct option is Option C.